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Answer :
Given data:
Mass of water:
[tex]m=1.70\text{ kg}[/tex]Initial temperature of water:
[tex]T_i=23^{\circ}C[/tex]Energy required to boil water is given as,
[tex]Q=mC_p(T_b-T_i)[/tex]Here, Cp is the specific heat capactity of water, and Tb is the boiling temperature of water.
Substituting all known values,
[tex]\begin{gathered} Q=(1.70\text{ kg})\times(4.186\text{ J/g}^{\circ}C)\times(100^{\circ}C-23^{\circ}C) \\ =(1.70\times10^3\text{ g})\times(4.186\text{ J/g}^{\circ}C)\times(100^{\circ}C-23^{\circ}C) \\ =547947.4\text{ J} \end{gathered}[/tex]Converting Joules to cal;
[tex]\begin{gathered} Q=(547947.4\text{ J})\times(\frac{0.239006\text{ cal}}{1\text{ J}}) \\ =130962.71\text{ cal} \end{gathered}[/tex]Therefore, 130962.71 cal of energy required to completely boil 1.70 kg of water at 23°C.
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Rewritten by : Jeany
The amount of energy is 1048.9 kcal to completely boil 1.70 kg of water that starts at 23°C.
To calculate the energy in kilocalories (kcal) required to boil 1.70 kg of water from 23°C, you need to consider the specific heat of water and the heat of vaporization. The specific heat of water is approximately 1 kcal/kg°C. To heat the water to boiling point (100°C), you first raise its temperature from 23°C to 100°C.
The energy required to heat the water to boiling is:
(100°C - 23°C) × 1.70 kg × 1 kcal/kg°C = 130.9 kcal
Once the water is at boiling point, you must also account for the heat of vaporization, which is approximately 540 kcal/kg. The energy required to boil the water is:
1.70 kg × 540 kcal/kg = 918 kcal
Adding these two amounts gives the total energy needed to completely boil the water:
130.9 kcal + 918 kcal = 1048.9 kcal
Therefore, it takes approximately 1048.9 kcal to completely boil 1.70 kg of water at 23°C.
